HDU 5742 It's All In The Mind（水题【多校联合】）

最新推荐文章于 2019-03-26 10:50:14 发布

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本文探讨了一个关于非递增数列的最大值比问题。教授张有一系列数，部分已知，部分未知，但知道数列是非递增且各元素总和不为零。目标是找到所有可能数列中a1+a2/Σai的最大值。

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http://acm.hdu.edu.cn/showproblem.php?pid=5742
It’s All In The Mind

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 68 Accepted Submission(s): 36

Problem Description
Professor Zhang has a number sequence a1,a2,…,an. However, the sequence is not complete and some elements are missing. Fortunately, Professor Zhang remembers some properties of the sequence:

For every i∈{1,2,…,n}, 0≤ai≤100.
The sequence is non-increasing, i.e. a1≥a2≥…≥an.
The sum of all elements in the sequence is not zero.

Professor Zhang wants to know the maximum value of a1+a2∑ni=1ai among all the possible sequences.

Input
There are multiple test cases. The first line of input contains an integer T, indicating the number of test cases. For each test case:

The first contains two integers n and m (2≤n≤100,0≤m≤n) – the length of the sequence and the number of known elements.

In the next m lines, each contains two integers xi and yi (1≤xi≤n,0≤yi≤100,xi

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>

using namespace std;

int gcd(int a,int b)
{
    if(b==0) return a;
    else return gcd(b,a%b);
}
int BB[100];
int main()
{
    int t,a,b,n,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&n,&m);
        int flag1=0;
        int A=0;
        int B=0;
        int flag2=0;
        int flag3=0;
        memset(BB,-1,sizeof(BB));
        for(int i=1;i<=m; i++)
        {
            scanf("%d%d",&a,&b);
            BB[a]=b;
        }
        if(BB[1]==-1&&BB[2]==-1)
        {
            A=200;
        }
        else if(BB[1]!=-1&&BB[2]==-1)
        {
            A=2*BB[1];
        }
        else if(BB[1]==-1&&BB[2]!=-1)
        {
            A=100+BB[2];
        }
        else
        {
            A=BB[1]+BB[2];
        }
        int p=2;
        for(int i=3;i<=100;i++)
        {
            if(BB[i]!=-1)
            {
                B+=BB[i]*(i-p);
                p=i;
            }
        }
        B+=A;
        if(A==0&&B==0)
            B+=1;
        int k=gcd(A,B);
        printf("%d/%d\n",A/k,B/k);
    }
    return 0;
}